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A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.510 m/s. The total mass of the sled, man, and rock is 97.5 kg. The mass of the rock is 0.320 kg and the man can throw it with a speed of 16.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled (in m/s) if the man throws the rock forward (i.e., in the direction the sled is moving). m/s Determine the speed of the sled (in m/s) if the man throws the rock directly backward.

Respuesta :

Answer:

a) v =  [tex]v = 4.70\times 10^{-3} m/s[/tex]

b)V = 0.059 m/s

Explanation:

given data:

speed of frozen lake is 0.510 m/s

mass of rock + man  is 97.5 kg

mass of rock is 0.320 kg

speed of throwing rock is 16.5 kg

By conservation of momentum...

(m + M)(v) = mv + Mv

(97.5)(0.510) = (0.320)(16.5) + (97.5 - 0.320)(v)

[tex]v = 4.70\times 10^{-3} m/s[/tex]

for backward..

(m + M)(v) = mv + Mv

[tex](97.5)(4.70\times 10^{-3}) = -(.32)(16.5) + (97.5 - .32)(v)[/tex]

V = 0.059 m/s

boffeemadrid

Answer:

0.45734 m/s

0.56601 m/s

Explanation:

M+m = Mass of sled, man, and rock = 97.5 kg

m = Mass of rock = 0.320 kg

M = Mass of Man and sled = 97.5 - 0.320 = 97.18 kg

[tex]u_1[/tex] = Velocity of rock = 16.5 m/s

[tex]u_2[/tex] = Velocity of sled

Here the momentum of the system is conserved

[tex](m+M)v=mu_1+Mu_2\\\Rightarrow u_2=\frac{(m+M)v-mu_1}{M}\\\Rightarrow u_2=\frac{(97.5)0.510-0.320\times 16.5}{97.18}\\\Rightarrow u_2=0.45734\ m/s[/tex]

The speed of the sled is 0.45734 m/s

In the case of opposite direction the speed will become negative

[tex](m+M)v=mu_1+Mu_2\\\Rightarrow u_2=\frac{(m+M)v-mu_1}{M}\\\Rightarrow u_2=\frac{(97.5)0.510-(-0.320\times 16.5)}{97.18}\\\Rightarrow u_2=0.56601\ m/s[/tex]

The speed of the sled is 0.56601 m/s

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